Do you know any explicit constructions of families of irreducible finite dimensional representations of the three string braid group?

I will describe what I already know and then invite you to suggest alternatives. This is somewhat open-ended so I will also pose some more precise questions.

This problem is solved for dimension at most five in
http://arxiv.org/abs/math/9912013 by Imre Tuba, Hans Wenzl
and published in Pacific Journal of Mathematics. So I am particularly interested in six dimensional representations.

The usual presentation is generators $\sigma_1$, $\sigma_2$ and the relation $\sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2$. An alternative is generators $s$, $t$ and the relation $s^2=t^3$ where $s=\sigma_1\sigma_2\sigma_1$ and $t=\sigma_1\sigma_2$. This shows we have a central extension of the free product of $C_2$ and $C_3$.

Given a representation of dimension $N$ take the dimensions of the eigenspaces of $s$ and $t$. Then we have $n_1+n_2=N$ and $m_1+m_2+m_3=N$. If the representation is irreducible then $n_i\ge m_j$. This data labels the irreducible components of the variety of irreducible representations of dimension $N$. The dimension of the component is $N^2-n_1^2-n_2^2-m_1^2-m_2^2-m_3^2+2$. For $N=6$ we have

(4,2) and (2,2,2) of dimension 6

(3,3) and (2,2,2) of dimension 8

(3,3) and (3,2,1) of dimension 6

(3,3) and (3,3,0) of dimension 2

The case (3,3) and (3,2,1) is interesting because these representations do not deform to representations of the symmetric group.

Can you give a representation and a braid $\alpha$ such that the trace of $\alpha$ and the trace of the braid given by reading $\alpha$ backwards are different? I have an example of dimension 12.

The description as a central extension of the free product means we have a construction which associates a representation to each invertible matrix (for each component).

I would like to improve on this in two ways. First this is unwieldy. Second I would like to be able to specify the eigenvalues of $\sigma_1$. This means passing to a covering of the moduli space.

1 Answer
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Bruce,
I can give you representants for an open piece of the moduli space (3,3) resp. (3,2,1). I'll add them in Mathematica-form so that you can plug them into any braid you like to work on. The variables a,b,d,x,y,z are the coordinates of the moduli space and l stands for a third root of unity (i dont know how to tell Mathematica to work with roots of unity, so if you know replace l by that thing).
I obtained these representants from the hexagon of 1-dml simples of the modular group and taking the dimension vector (1,1,1,2,1,0) as required. The corresponding moduli problem is then rational (in x,y,z) over that of pairs of 2x2 matrices of rank one. these are rational (in a,b,d). Hope this helps (and sorry for the lengthy formulas below).

EDIT March 9th : The component (3,3;3,2,1) is not able to detect braid-reversion. On the other hand, the component of 6-dimensional representations (3,3;2,2,2) can. A minimal braid that can be separated by this family from its reversed braid is s1^-1s2^2s1^-1s2. One can show that every irreducible component of simple B(3)-representations has a Zariski dense family parametrized by a minimal rational variety. For representations of dimension <= 11 explicit such families are given in this note.

Each of these families can then be turned into a family of 3-string braid invariants over Q(rho) for rho a primitive 3-rd root of unity by specializing the parameters to random integers. For the family (3,3;2,2,2) mentioned above this can be done in SAGE as follows :

The following tests should give non-zero invariants : 6.3,7.5,8.7,8.9,8.10 (which are known as 'flypes' in the theory) and 8.17 (the non-invertible knot with minimal number of crossings). I tried to include all details in the note Rationality and dense families of B(3) representations. All comments to this text are welcome. I like to thank Bruce Westbury for providing me with feedback.

I'm impressed. I did try that but I got something much more complicated.
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Bruce WestburyFeb 17 '10 at 17:10

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Hi Bruce, I'm glad you liked it! Btw. it was good timing posting this so early, I was reading my morning paper when I glanced at MO and saw your question. I scribbled a bit on the newspaper and decided I wouldnt be able to do the calculations by hand so I cycled over icy paths to the university to access Mathematica. It took me a while to find the braid, but it sure was a morning well spend! All the best :: lieven.
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lieven lebruynFeb 17 '10 at 21:46

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The note referred to at the end gives a very thorough answer to the problem. It does more than I hoped for.
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Bruce WestburyMar 9 '10 at 8:46